# What is the average speed of an object that is moving at 4 m/s at t=0 and accelerates at a rate of a(t) =2-t on t in [0,3]?

May 16, 2017

$5.5 \frac{m}{s}$

#### Explanation:

I'm assuming this is related to one-dimensional motion. Combining the integral functions for velocity and position, the equation for position with respect to time is represented by

$x = {x}_{0} + {\int}_{0}^{t} \left[{v}_{0 x} + {\int}_{0}^{t} {a}_{x} \mathrm{dt}\right] \mathrm{dt}$

So, since the initial velocity ${v}_{0 x} = 4 \frac{m}{s}$, the postion equation with respect to time from this is

$x = 4 \frac{m}{s} \left(t\right) + 1 \frac{m}{{s}^{2}} {\left(t\right)}^{2} - \frac{1}{6} \frac{m}{{s}^{3}} {\left(t\right)}^{3}$

and thus the position of the object at time $t = 3$ is

$x = 4 \frac{m}{s} \left(3\right) + 1 \frac{m}{{s}^{2}} {\left(3\right)}^{2} - \frac{1}{6} \frac{m}{{s}^{3}} {\left(3\right)}^{3} = 16.5 m$

The average velocity on the interval $t \in \left[0 , 3\right]$ is thus

${v}_{a v - x} = \frac{\Delta x}{\Delta t} = \frac{16.5 m}{3 s} = 5.5 \frac{m}{s}$